Show $\frac{n^2}{2^{\sqrt{\log(n)}}} \geq \frac{n}{2}. $ So we know that $\frac{n^2}{2} \geq \frac{n}{2}$, but I'm stuck proving that $\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$. Am I missing something?
 A: Let $\sqrt {\log n}=u,\thinspace u≥0$, then we have
$$\begin{align}&\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n}{2}\\
\iff &\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n^2}{2n}\\
\iff &n≥2^{\sqrt{\log(n)}-1}\\
\iff &e^{u^2}≥2^{u-1}\\
\iff &e^{u^2}>e^{u-1}>2^{u-1}.\end{align}$$
This means, the equality is impossible.
Therefore, we have
$$\frac{n^2}{2^{\sqrt{\log(n)}}} > \frac{n}{2}.$$
A: Notice that the expression $\frac{x^2}{2^\sqrt{\ln(x)}}$ is well-defined if and only if $x\geq1$. Division by $x$ then shows that your inequality is equivalent to
$$\frac{x}{2^\sqrt{\ln(x)}}\geq\frac{1}{2}$$
We can prove this inequality with calculus:
Consider the function $f:(1,\infty)\to\mathbb{R}$ defined by
$$f(x)=\frac{x}{2^\sqrt{\ln(x)}}$$
This function is differentiable, so any maxima or minima in its domain will have derivative $0$.
Using the quotient rule and the chain rule, we find that for every $x>1$, the derivative of $f$ at $x$ is
\begin{align}
f'(x) &= \frac{2^\sqrt{\ln(x)}-x\cdot2^\sqrt{\ln(x)}\ln(2)\cdot\frac{1}{2\sqrt{\ln(x)}}\cdot\frac{1}{x}}{\left(2^\sqrt{\ln(x)}\right)^2}\\
&= \frac{2^\sqrt{\ln(x)}\left(1-\frac{\ln(2)}{2\sqrt{\ln(x)}}\right)}{\left(2^\sqrt{\ln(x)}\right)^2}\\
&= \frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}
\end{align}
We now solve the equation $f'(x)=0$ to find $f$'s critical points, if any exist.
\begin{align}
f'(x)=0 &\iff 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}=0\\
&\iff \frac{\ln(2)}{2\sqrt{\ln(x)}}=1\\
&\iff \frac{\ln(2)}{2}=\sqrt{\ln(x)}\\
&\iff \ln(x)=\frac{\ln^2(2)}{4}\\
&\iff x=\exp\left(\frac{\ln^2(2)}{4}\right)
\end{align}
Thus, $e^{\frac{\ln^2(2)}{4}}$ is the only critical point of $f$. This actually corresponds to a minimum of $f$, which we can prove by showing that $f'(x)$ is negative for $1<x<e^{\frac{\ln^2(2)}{4}}$ and positive for $x>e^{\frac{\ln^2(2)}{4}}$.
\begin{align}
1<x<e^{\frac{\ln^2(2)}{4}} &\implies 0<\ln(x)<\frac{\ln^2(2)}{4}\\
&\implies 0<\sqrt{\ln(x)}<\frac{\ln(2)}{2}\\
&\implies \frac{\ln(2)}{2\sqrt{\ln(x)}}>1\\
&\implies 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}<0\\
&\implies f'(x)=\frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}<0\text{, since }2^\sqrt{\ln(x)}>0
\end{align}
This proves that $f'(x)$ is negative for $1<x<e^{\frac{\ln^2(2)}{4}}$.
\begin{align}
x>e^{\frac{\ln^2(2)}{4}} &\implies \ln(x)>\frac{\ln^2(2)}{4}>0\\
&\implies \sqrt{\ln(x)}>\frac{\ln(2)}{2}\\
&\implies \frac{\ln(2)}{2\sqrt{\ln(x)}}<1\\
&\implies 1-\frac{\ln(2)}{2\sqrt{\ln(x)}}>0\\
&\implies f'(x)=\frac{1-\frac{\ln(2)}{2\sqrt{\ln(x)}}}{2^\sqrt{\ln(x)}}>0\text{, since }2^\sqrt{\ln(x)}>0
\end{align}
This proves that $f'(x)$ is positive for $x>e^{\frac{\ln^2(2)}{4}}$. We infer that $f$ has an absolute minimum at $e^{\frac{\ln^2(2)}{4}}$, so for every $x>1$,
\begin{align}
f(x) &\geq f\left(e^{\frac{\ln^2(2)}{4}}\right)\\
&= \frac{e^{\frac{\ln^2(2)}{4}}}{2^\sqrt{\ln\left(e^{\frac{\ln^2(2)}{4}}\right)}}\\
&= \frac{e^{\frac{\ln(2)}{4}\cdot\ln(2)}}{2^\sqrt{\frac{\ln^2(2)}{4}}}\\
&= \frac{2^\frac{\ln(2)}{4}}{2^\frac{\ln(2)}{2}}\\
&= 2^{\frac{\ln(2)}{4}-\frac{\ln(2)}{2}}\\
&= 2^{-\frac{\ln(2)}{4}}
\end{align}
All that remains to be shown is that $2^{-\frac{\ln(2)}{4}}>\frac{1}{2}$:
\begin{align}
2^{-\frac{\ln(2)}{4}}>\frac{1}{2} &\iff e^{-\frac{\ln(2)}{4}\cdot\ln(2)}>\frac{1}{2}\\
&\iff -\frac{\ln(2)}{4}\cdot\ln(2)>\ln\left(\frac{1}{2}\right)\\
&\iff -\frac{\ln(2)}{4}\cdot\ln(2)>-\ln(2)\\
&\iff \frac{\ln(2)}{4}<1\\
&\iff \ln(2)<4
\end{align}
The last of these is true because $\ln(2)<1$, so we have that $2^{-\frac{\ln(2)}{4}}>\frac{1}{2}$. Thus, for every $x>1$,
$$f(x)>\frac{1}{2}$$
or
$$\frac{x}{2^\sqrt{\ln(x)}}>\frac{1}{2}$$
A: For all $t > 0$, we obviously have $t > \sqrt{t} - 1$. Therefore we also have
$$
e^t > 2^t > 2^{\sqrt{t} - 1},
$$
or equivalently,
$$
\frac{e^{2t}}{2^\sqrt{t}} > \frac{e^t}{2}.
$$
Now just let $t = \ln x$.
A: For $n\ge 1$ we have
$n^2/2^{\sqrt {\log n}}\ge n/2\iff$
$ n/2^{\sqrt {\log n}}\ge 1/2 \iff $
$ 2n\ge 2^{\sqrt {\log n}}\iff$
$\log 2 +\log n\ge (\sqrt {\log n}\,)\log 2\iff $
$\log 2 +\log n - (\sqrt {\log n})\log 2\ge 0 \iff$
$(\sqrt {\log n}-(\log 2)/2))^2+\log 2 -(\log 2)/2)^2\ge 0.$
